for randomly selected adults, IQ scores are normally distributed with a standard deviation of 15. the scores of 14 randomly selected college students are listed below. use a 0.10 significance level to test the claim that the standard deviation of IQ scores of college students is less than 15. round the smaple standard deviation to three decimal places.
115 128 107 109 116 124 135
127 115 104 118 126 129 133
To test the claim that the standard deviation of IQ scores of college students is less than 15, we will conduct a hypothesis test using a significance level of 0.10.
Let's go step by step on how to perform this test:
Step 1: State the null and alternative hypotheses.
The null hypothesis (H0): The standard deviation of IQ scores of college students is equal to or greater than 15.
The alternative hypothesis (Ha): The standard deviation of IQ scores of college students is less than 15.
Step 2: Calculate the sample standard deviation.
The given IQ scores of 14 randomly selected college students are:
115, 128, 107, 109, 116, 124, 135, 127, 115, 104, 118, 126, 129, 133.
To calculate the sample standard deviation, you can use a statistical software or a calculator. The rounded sample standard deviation is 9.927.
Step 3: Calculate the test statistic.
The test statistic for comparing the sample standard deviation to a known value is calculated using the chi-square distribution. In this case, since we are testing if the standard deviation is less than 15, we will use the chi-square distribution's lower tail.
The test statistic formula is:
chi_square = (n - 1) * S^2 / sigma^2
Where:
n = sample size
S = sample standard deviation
sigma = hypothesized standard deviation (15 in this case)
In our scenario, n = 14, S = 9.927, and sigma = 15.
Calculating the test statistic:
chi_square = (14 - 1) * (9.927^2) / (15^2)
chi_square ≈ 6.036
Step 4: Determine the critical value.
Since we have a one-tailed test, and the significance level is 0.10, we need to find the critical value from the chi-square distribution table.
With 13 degrees of freedom (n - 1), the critical value from the table for a significance level of 0.10 is approximately 7.815.
Step 5: Compare the test statistic with the critical value.
If the test statistic is less than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In our case, the test statistic (6.036) is less than the critical value (7.815).
Step 6: Make a conclusion.
Since the test statistic is less than the critical value, we have insufficient evidence to reject the null hypothesis. Therefore, we fail to reject the claim that the standard deviation of IQ scores for college students is less than 15 at the 0.10 significance level.
In simpler terms, based on the given data, we don't have enough evidence to conclude that the standard deviation of IQ scores for college students is less than 15.